 Okay. Let us begin. I was showing you how loads dropped out in sets of bolts. Here's some tests they did. P on the middle plate, P over 2 on the outside plate. This obviously depends a lot on how stiff the two plates are and the material out of which they are made, but basically you can cover any of that variation. Here is what they found. Here is the middle plate pulling. You will notice that the hole in which it was put really shows up here. This bolt is really taking out load and dumping it into the next plate. Here the next one is not bent as much. This one is not bent as much. This looks like it has still to go and by the time this one gets really deformed, this one will be even more deformed, but it will still be holding its load. It will just be severely deformed. We discussed block shear. We mentioned that we don't even have a symbol for F sub u for shear. Since it is 6 tenths of F sub u ultimate in tension, we just use, we don't even have that symbol, you know, just to make sure you know what I'm talking about. It's F sub u in the book, 6 tenths of F sub u. If you ever have something in shear, a surface in shear, then it will be 6 tenths of the tensile strength. Now, if the bolt people use that or not, that's the bolt people's problem. And if we use their specifications in our specs, then, you know, we'll do whatever they do. But for the types of steel we use, structural steel, A 36, all that kind of stuff, we're going to be using 6 tenths of the ultimate for the ultimate in shear and the yield in shear will be 6 tenths of the yield in tension. We mentioned that as the steel's got stronger, we started having problems with the little corners breaking off. When we would pull these little corners out, we would notice that some of it got sheared and therefore you really ought to be applying F sub ultimate in shear, which would be 6 tenths of F sub ultimate in tension, which would be 6 tenths of F sub y on the sides that are sheared, wiped across. And the little end here that's actually is placed in tension, you ought to be using F sub u, T or just F sub u on that area to determine how much load it took to pull this corner off. That's called block shear. Here's a typical channel or a wide flange where someone's bolted something in the middle. I can't tell if it's a channel or a wide flange. And it tore this little block out of the middle and that little piece of steel that used to be called the web of the channel is now part of the plate. On the other hand, it could be that the channel is stronger or thicker than the gusset plate in which case the little plug will come out of the gusset plate. If they're both made out of the same steel and the channel is thicker than the gusset plate, then I know immediately to go check on the gusset plate because the pattern of failure is the same and I would use the thinner of the two materials. If this is made out of a high strength material and that's a low strength material and one is thicker and one is thinner, I wouldn't have any idea. I'd just have to check on both. Here are the parameters that you will be using. This happens to be a channel, flange, web, flange. Happens to be a 6 by 13, about 6 inches deep. It's not about it's 13 pounds of foot pretty much right on the button. They have 7-8 inch bolts in the connection which means these are, as far as you're concerned, 1 inch holes. The pattern, that's the spacing, spacing, spacing, that's the gauge. There aren't any slanted lines there so it really won't matter whether or not you know what the gauge or the spacing is because it won't show up in the equation. In the thickness of that web, check it out in the book, .437 inches. This little block will pull out and the first thing you'll have to do is calculate the gross areas. On the end, I'll tell you that the gross area and tension is the thickness of the channel times the 5 inch dimension across the bolt holes and in shear, there are two shearing surfaces, one on the top, one on the bottom. That's the area of gross tension. Here's the area of gross shear over two. The dimension between the last hole and the entire block being pulled out was 8 inches so that surface area is 8 by .437 and there are two sides. In the calculations they ask you to perform for strength, you will find you need area gross tension and area gross shear. There's your numbers. Since there are holes in here, you will have to take the effect of these holes out of the gross area to get the net area. To get the net area, you would take the 8 inch length, it was 8 inches from here to here. You would subtract one 1 inch hole, two 2 inch holes, two and a half 1 inch holes because this last one only had a length of a half a hole, a radius big. That would be multiplied times two and a half is multiplied times one to give you the length lost along that line. 8 minus the length lost gives you the net length. Multiply the length of that line times the thickness of the web and there's two of them so your net shear area is this number right here, whatever that calculates. Then on the tension end of things, you've also lost a half of a hole and a half of a hole on your net tension area. It used to be 5 times .437. Here you've lost two holes with two holes that's a half of a hole and a half of a hole, each one of them was a half inch across, half inch across, half inch across and that length times that distance gives you the cross sectional tension area on the back side here with the holes on each side. Now all we've got to do is put all those pieces together like the specs want you to do to determine how strong they are. I'll tell you this, it makes sense. If all of a sudden they go out and test it and something doesn't make sense, the first thing they do is they change the specs until it's true. Then they try like crazy to explain why it's true and if they can never figure it out, they don't care. I mean they care, but they don't know. If they say look, multiply that's .5, you have to do it because test is shown that's what you need to do. Now this is a typical failure pattern where there was a plate on the top of a wide flange and a plate on the bottom of a wide flange or possibly a S and it failed. It failed on this shear surface, two, three, failed on four shear surfaces and it failed on these relatively small tension areas. So your area net shear would be the length from here to here minus a half, one and a half, two and a half, minus two and a half holes times the thickness. The net tension area would be how far it is they drill the hole from there to the edge minus a half a hole. That'll give you this dimension times what is that thickness of? Thickness of the flange, that's correct. And there are four of those. So this right here pointed to that's a fourth of the area net tension. That's the way these things really fail. This is not how they fail. I mean, you know, it could be, you could say why not? You know, it's torn loose just like the other one had pieces torn off of it. Well, if you'll notice you have the same shear here as you have shear area here. So the shear area on both thoughts of failure are the same. The difference is the tension area is small plus small plus small plus small times that's small times four. That's medium. Look at the tension area here. Very large, massive, massive. Here's some discussion and that's okay. You see something wrong in here? Let me know. No, that's fine. And talking to the people like that, that's fine. And I don't ever fuss about that. The real reason I fuss is because I thought you had a complaint. No, but I'm waiting for somebody who has a complaint. Well, okay, you're not, you're in other words doing ugly things to my drafting skills. I was in a hurry. The shear is listed. Well, it is over four. See, there's one, two. The tension in the shear are badly labeled. This is not a tension area. This is a shear area. That's what I thought that you were discussing over there and I wanted to go ahead and get that from you. But that's that's not a problem. Well, you know, the first time I wrote it, I didn't catch it either. But last semester, I can assure you a lot of people told me about it. Then I thought, well, that was pretty nifty. Let's see how many you'll catch it this time. And you did, you will. The shear areas aside, that's been wiped, the tension areas, the one that's been pulled straight out in tension. But the thing, the reason this can't control is there's just too much extra tension area with the same shear. There's nothing wrong with checking this. It's not a good idea to check it on an exam because you're not going to get, you will get the right answer for if this failed like this, which it can't do. And then when I write down something less than 100%, I'd say, well, I would have, but I can't do because they can't do this. Same way with an angle. And then I get some people who think the angle fails across here like that. The real way it fails is across the same shear area with a little bit of tension area out on the end. So the bottom line is on all of these things, the shear line, the failure line plane never goes across outstanding legs or elements or whatever you want to call them. You don't quite understand the words there will then just draw the two potential failure things and take the one that has the least tension area. Sheer areas usually the same. Yes, sir. Yeah, I do. Do you like to buy one? No. Well, sorry, your fees only get poorly drawn hand things. That's all you get. Now, here is a beam that is being framed into a girder. And they've had to cut the top out of it so that the floor will remain at the same elevation on the whole thing and they pull concrete across the top. They'll come in here and they may like on this side and show you one of them show that in three days that goes like that. It goes back like that. That's why I don't do these goes like that goes like that goes like that goes like that goes like that. There we go. That's getting close. And then it's welded down in here like that. And it's a plate and it's welded onto the girder or it could just be an angle that's bolted on and it'll have some holes in it like this. What happens is when you both these two together, the load on top of the beam has a tendency to tear this little corner out. If you still have that top flange, you won't get a failure. You'll put so much extra steel up there that it won't happen. So this is only when you cope a beam, then you have a tendency to block shear, rupture this little piece out of the web of the beam. And the stresses they find are nicely and uniformly distributed. And you get the full strength of this little tension piece down here. If on the other hand, you have two columns of bolts, sometimes three columns of bolts because you have a whole lot of load. It turns out the stresses are not nicely uniformly distributed. They are they look more like this. They have found that you do not get the full strength of this little block in tension around the bottom because of their not being uniformly distributed down at the bottom. And they make you apply a factor of point five to the tension term. That's the only one that they list that you have to do that to. All right, now back to the problem at hand, we'll have a nominal strength of shear of six tenths tension ultimate, that's shear ultimate times the net shear area that stress times area makes sense. And on the tension end of things, you'll have the full no point six here because there's no shear involved times the steel that's really left in tension net means left in shear net means left in tension after the holes are drilled out and you have accounted for them in your calculations. So the strength according to the specifications, you get the allowed of the allowed ultimate strength of the metal in shear times how much shear area is really there plus the ultimate intention times the area that's really still there in tension. The only thing wrong with this thing is what they find is every now and then this tension term actually doesn't have a nice uniform stress under it. And therefore you have to use a use of bs of point five us bs it's on this page. I got it on page 67 a I'll show it to you. So generally speaking, your equation reads shear ultimate times how much areas in shear plus a correction under certain conditions times attention term ultimate intention times how much steel is really there. Except sometimes they find that your connections are so long that they shear and they deform a whole lot. There was a really do deform and they deform so much that the little tension guy down there, he is doing this. He coming coming up in tension. He's yielding intention. Everything is fine. But he's getting so much deformation. He's actually getting out in here. And they say that's not a good idea. And so what they do is they say, if you go with the gross area up to the yield stress in shear, then we want you to stop. In other words, you can apply the ultimate stress to the steel that's really there. But we found that if you exceed f sub y in shear times the gross area, it's deformed so much we don't want you to continue. You must stop. So first thing you do is you calculate your number here. This number will be the same in both cases. But if this term right here has driven that up above f sub y, six tenths f sub y gross have to stop. So calculate this, then calculate this. And I don't know it's always interesting, you know, make sure this is less than that. That tells you what you're supposed to do. The truth is it could just as easy as said, make sure this is less than that. It works either way. You have two terms and you want to take the lower of the two predictions. When you get through calculating the nominal resistances, because it is a breaking or a rupture situation, we want you to multiply it times 0.75. Notes. This is part 16, chapter j, section 4.3, but the page, oh yeah. I don't remember. It looks, it seemed like the, that's, I was just trying to make sure you know these page section things again. In other words, this thing right here is in chapter j. It's in section 4.3, but they just call it section 4.3. And though it's really chapter j, they call it just section j3. So that's a little bit different from what it's really, what it really is. It's in chapter j. Got an example. The terms you are going to need to work this, you're going to need the f sub u intention for your steel in use. You're going to have to know the net shear in shear, the net area in shear. You're going to have to know the net area intention. You're going to have to know the ultimate. You're going to have to know if the stress is actually on a beam, on the end of a beam, where you have more than one column of bolts. This you'll already, this you added, you have to know this. And you'll have to, you already calculated the gross area, so you could calculate the net area from it, so hang on to that number. And then everything else here is the same. Got an example. He has 7 eighths inch bolts, so he's got one inch diameter holes, a 36 steel. You get these dimensions from page one dash 48. That's where we put bolts in angle legs most of the time. I wouldn't guarantee you a homework problem has that, but generally speaking that's considered good practice. They're placed at three, three and one and a half. That tells you the length of the shear block and that tells you the width of the shear block. You'll be calculating gross shear and gross tension. Then you'll take out the holes. You'll take out that length times that thickness. You'll take out that length times that thickness. You'll take in this half of a hole times that thickness to turn the gross shear into the net shear. Now we'll get to that in a minute. Here are the specs. Strength of elements in shear, block shear strength. Available block shear. Here's your first equation. Here is if you get that shear area so that it's so highly stressed that it actually may tend to break the thing on the end, they don't want you to go that high. They want you to limit it to 6 tenths f sub y area gross shear. Calculate both terms, see which is a smaller. Then apply a resistance factor of 0.75 and he's got little comments. Very interesting. Well that's where I just showed you. Typical cases where this should be taking this 0.5 are illustrated in the commentary. It'd be nice if you'd tell me where but you know after a while you learn these things it's on that page and I got a copy of it. We already looked at it. Maybe not 67a. Back to the problem at hand. Shear areas are seven and a half inches long by three eighths inch thick. There are two and a half hole diameters each an inch in size. So we take the net shear area as three eighths times seven and a half. That's the total length. Subtract two hole diameters to get us the net shear. Net tension area is the area one and a half times three eighths minus a half a hole times the thickness. Three eighths there's one and a half minus a half of a hole. That much steel. There's the factor of 0.5 is used because there's a half a hole. I thought he's going to put a UBS on me here. I said I don't see that. 0.5 is there's only a half a hole missing in this dimension right here. Since the block shear occurs in an angle UBS is equal to one. Why? Because of 16.1-129 equation j page 66a. Here is the equation. Here's the page it's on. Here we got 87 kips including everything in the first half of our requirements. It's supposed to make sure it does not exceed this number. Extensive 36 where did the 36 come from? 36 steel that's right times the area gross in area gross in shear. You'll notice you have a reduced stress but you have an increased area. Then the second term remains the same it's 82.51. Look at that. This thing evidently would deform so much in shear that he's not going to let you let it go all the way up there. He's going to restrict you to 36 times 0.6 times the gross area. F sub y versus F sub u area gross in shear versus area net in shear. So you get the lower of the two 82.51 kips. Then to go to the design you got a multiple times the resistance factor of 0.75. When you bring up your 1.4 dead and you bring up your 1.2 dead plus 1.6 live plus 0.5 snow and you put in your earthquake and you put in your wind and you put in all that stuff do not come up here with a number bigger than 61.9. I won't let you through the door. You say I've got to have a bigger number then you're going to have to pay for another design because that angle didn't work. Yes sir. UBS is U for block shear when the block shear element is not uniformly distributed across the bottom. Here it is right here. It's on page 16.1-412. That may in our notes. Here is a little block that is pulled out only if the beam is coped. If it's not coped you don't have to check block shear. Here it tore out a corner of an angle. As you might imagine this is the tension side and therefore that's where the stresses are problematic. Here's an angle that's bolted on or welded on and the stresses are uniformly distributed. Here's a welded angle uniform. Here's a welded angle on a coped beam. Here is a bolted plate or angle to the column. One row one column of bolts. Here's an angle where the sides torn off rather than the middle out of the gusset plate. Here is a plate a gusset plate with bolts on it and those are considered uniform and then all of a sudden you run across one. Here's your column. Here's your beam. Here you have two columns. They have found the stresses are not nicely distributed. They do get higher than anybody thought they would because of that and therefore cases for which use block shear is .5. If you have three columns use a .5. Yes sir. Well you get a lot of gain because when you put the bolts on here the bolts on here were only half as strong as we needed. So the bolt guy did that and she came in with two rows of bolts and I said oh man you're killing my beam. She says that's not my problem. She says if you'd like you can get a beam where I can get eight bolts in a column there. So man that would be a 48 by 970 beam. She says not my problem. She says all I know is I need eight bolts. So I put them that way then I got to go find somebody who can do this job with the BS box shear U down to .5. Well with what happens is the load comes in here and it goes through the bolts tending to tear this corner out. The bolts press against the angle. The angle then presses against the column and of course we move on from there. The column load goes on down into the footing and it goes on down into the earth and that's the end of it. Yes sir. Yeah they they definitely tend to shear a block out of the out of the gusset plate and they also tend to shear uh they won't like this one right here won't have a block shear on the angle no if that's your question. There'll be no block shear on this one on the angle. There will be a block shear in the in the gusset plate. In other words it'll pull a little chunk right out of there. All right another thing we worry about is a radius of gyration. You can put plates all over the place you shouldn't you should try not to make them so thin that they have a tendency to rattle around when the wind blows. In other words if you find that this plate works nicely you need to really check its radius of gyration about its minor axis to see what's going on because if its L over R is very large then you get these ghost noises. The wind blows and you hear clank clank clank clank clank and the guy in the office says what the devil is that? Then we drill through the hole through the wall and we look around and we see this thing is flopping around. Plenty strong no problem but then you're gonna have to brace it somehow so it doesn't hit the other rod or something. It is suggested that the slenderness ratio of whatever you use be it an angle or a plate or you name it shouldn't be over 300 but it is a suggestion. It's a notification that if you make it much over that and don't make some special provision so you don't have some of these things slapping into each other you're going to have some noises that you don't want. Now looking at it from another point of view if you tell me you're designing attention member then p sub u should be less than the nominal strength times of an appropriate reduction factor a resistance factor turned around the other way. Phi being 0.9 for yield then p nominal would be f yield area gross. That's the number of course that you're trying to get larger than p sub u. Your ultimate request. You're looking for if you know the steel you're going to make it out of if you say it's going to be an angle well okay I'm going a 36 then I can tell you you better go find me an angle that has a gross area greater than 1.2 dead plus 1.6 live or 1.4 dead or whatever divided by 0.9 divided by f sub y. So if I'm going to design something that's pretty nice to know. Otherwise right off the bat I'll take your load divided by 0.9 to make the load bigger divided by f sub y to get p over a a p over sigma to get a and then that will be I'll have to find an area on the angle that's got that much if you don't give me one that big to begin with I can't even get started. To avoid fracture you want the same idea you want the fracture resistance factor 0.75 times f ultimate times a effective area you want it greater than the requested load so the effective area ought to be greater than this number and if you don't at least start out with that again I don't have a chance now you may say well that's not going to be as easy because the effective area depends on the length of the connection and it depends on all kinds of things I say I agree but at least it's a start hopefully it gives you something to start with centerness ratio will be satisfied if the radius of gyration is greater than l over 300 if you're talking about a wide flange you're going to use this about that axis for the radius of gyration if you're talking about an angle be talking about this axis because that is the minor axis in other words once you put it on a plate you can force it to bend about this axis but once it's out between the two ends it's going to buckle or going to bend or going to flop around about the z axis so you'll be looking for a radius of gyration about the the z axis. This has got an example got a tension member 5 foot 9 long service load dead load 18 live load 52 he wants to remember with the rectangular cross section shouldn't be much to this don't mean to just blow through them but day 36 steel he wants the connections in a single line he wants seven eight inch diameter bolts which gives you one inch diameter holes there's your gross area cut across here there's your net area there's your load 104.8 piece of view here's your required gross from our equations we just did piece of view nine tenths f sub y piece of view nine tenths right out of the table for a 36 steel so do you use a 36 that many square inches in the plate the effective area has to be piece of view of 0.75 f sub u we need that much effective area now in this case not going to be quite as hard because the effective area is the same as the net area because we don't have any unconnected outstanding elements now years of experience going that and i know that and i'm admitting that he just says try an inch well we mean try an inch why not try six inches well because it's a horrible thing i said well why not try half an inch i said well it's not good so you just got to start you and me we just got try an inch therefore the gross width would be the required gross area divided by how thick you're trying that means has to be 3.235 inches now here's the point if you find this is a that a one inch thick plate requires a really wide plate you may realize that it's going to have a very low radius of gyration and you may say i don't i don't think a one inch plate is going to be a real good plate or if you find after all is said done that thinking the plate should be one inch thick and then it only has to be a little bitty like this you know he said i don't think i've ever seen a plate where they drilled it through the edge like that and so that's not a good choice and all you can do is just rummage around in there until you get something that looks reasonable that's about it turns out this one works out all right one by three and a half is reasonable if it was a one by one that wouldn't be good now then we're going to check the effective area on this trial cross section so we go one by three point five four gross we subtract out the area of a hole that's the diameter of the hole that's the thickness of the plate and that's able to two and a half square inches which turns out more than we needed so that's pretty good plate and anytime anything comes out that close you can be sure somebody already tweaked the numbers before he wrote it down in the book here is your bh cubed over 12 for the moment of inertia of a one by three and a half plate here's the area i'm sure you remember that the equation for r is a square root of i over a or have access to it calculate the uh minimum radius of gyration i don't know if your book has inches squared anybody got a book well anyway if it does mine does it's not inches squared a radius is a radius it's a length so that's a minor typo maximum l over r 239 less than suggested not required by the specs it's okay use that plate i was curious anybody using the ta raise your hand what he's using the ta we got down there for 20 hours a week poor guy must really be bored okay read this uh example illustrates it wants to require there's determined procedures no no it doesn't this is a procedure same but same for l or fd and ast we don't do ast so we have no idea uh less than eight inches wide so it's a plate instead of a bar i don't care about that here he's saying for the first time you and i've already been using it where these gauge lengths are on angles on page one dash 48 in the specs here he's going to use one he's got an angle got 35 kips dead and live with three-quarter inch diameter bolts means you and i are going to be using seven eight inch holes factored load 154 kips required gross area from our equation same as we did last problem 4.75 square inches effective area has to be this much don't know how i'm gonna get that effective area to tell you the truth because i got a lot going on here and as best i know no he has he told me the length of the connection i can check that uh the radius of gyration should be uh l over 300 for the radius of gyration so it calculates we ought to have this number on our angle when we pick it find the lightest shape says we search the dimensions and properties tables first off he wants an unequal leg angle i have no idea why let me go ask him he says well on all the equal leg angles the angle sticks up pretty far and i and i lose a lot of strength because it's a five by sticking the air five angle he says if you get me an unequal leg angle then you'll probably find me an eight by four angle that'll that'll work and invariably you'll lay the long leg flat on the plate and you'll save me some money so i want an unequal leg angle gotcha uh that has the smallest acceptable gross area and check it's effective net area so about all we can do is pull one that looks like it might work on gross area and then see if we can really make this work uh starting at either end of the table here's the table the properties table we want something that has what area do we need 4.75 gross so we go for 4.75 that one might work six by four by half that one might work that's got an area over 4.75 no no no forget it that's a five by five that one might work that's a five by three and a half by five eights that one might work but it's an equal leg angle so got a couple of choices here we can try yes sir oh yeah first place don't choose one this lower i mean if you pick one this lower you're doomed already now then should you pick this one or this one you know that's that's already chancey but i mean you know we don't know it's possible that'll work so i'm going to give it a try you say well it's probably not going to work so i think i'll just try this one and then i do that one and it works and then you spend an extra 170 thousand dollars in steel and i got the bid and you're living under a bridge i'm going to try that one you know and if it won't work then i'll suck it up and go to this one knowing you had to do the same thing obviously you know in a class like this you don't want to use a whole bunch of time doing things repetitively in the real world like i say get paid by the hour and you have great you know fun when you really find out you saved a hundred thousand dollars in steel by knowing what you're doing uh here are the other dimensions here's the radius of gyration for the angles if we decided to pick them to check if the l over r is okay here is x bar if we are going to get ready to use a uh yeah a u not a u but a one minus x bar over l term forget if that term had a name to it here's some more angles oh that's bad it sounds like this angle didn't work go so we're going to try the six by four by half was the one that was pretty close its personal net area would be area gross minus area the holes there's the area gross there's the area the holes we lost two uh holes of that diameter times the thickness of a half inch angle left us with 3.875 says look you don't even know the length of the connection so you show them where x bar was was no help at all and i understand that therefore what i'll do is i will go to a table that gives me acceptable values whether or not i have the length or not is this table right here these are our shear lag factors here are angles right here oh excuse me forget that we don't know the l and therefore we're going to go down here to single and double angles four or more fasteners 0.8 that sounds good that sounds good where's the picture of that thing i forget how many boats it had four or more it's got four more boats if you only had three i'd be stuck with a 0.6 therefore having four or more i get 0.8 multiplied in there here is my uh net area is this much permitted to use 0.8 whether or not i come out with something even smaller or even worse than this later on this is still permitted but i'll we'll check it for it's over with i know i won't i'm gonna bother checking anything i'm already down to 3.07 square inches and i'm supposed to have 3.54 3.54 for the effective so i'm gonna have to use a different angle you go through the same calculations again the calculations must have been on the bottom of that page they are there's not any more to it than that here's our 5.8 for the gross here's the lost area 4.9 4.9 times 0.8 3.94 is bigger than we need it's okay then he checks the l over r satisfies everything all right now there are tables that will help you design they are really rough by that i mean they're just to put you in the ballpark is all they are going to do first off is not a problem to go get somebody with the right gross area but that effective area is a problem so what they have done is they have put together some tables like so the table is for the kind of steel you're going to use for that kind of a item and here's the shape 5 6 by 4 by half inch angle tells you the gross area and it tells you how strong the thing is in yielding so if you need 154 kips this is this guy will do it question is is how about the effective area on the thing and without a doubt using the 0.8 or something like that he didn't tell us and so you're just going to have to you have to check this when you're through he says if you can live with the effective area being 0.75 times the gross that includes the you that includes the holes that includes everything that doesn't include blocks here you still got to check that too i said well i don't i don't accept that's anywhere near right i have no idea he says well then don't use my tables go away well so wait wait wait i guess i'm not that firm on this point in other words he says that's not a bad average it's probably half of them are too big and half of them are too small and but it'll give you 155 kips it gives you a point to start immediately and then of course you don't have to check the gross area because you know it's got the gross area capacity all you got to do is go check all of the other little doodads in there all right see you next time go through the example he has on that if you want to use those design tables they can be handy thank you the picture you asked me about this is the kind see how the middle piece is just almost destroyed and the back part is broken now had they kept on pulling it out they'd have pulled us out across the shear as well as this then i could have shown you one that was fully broken but do i have one where the channel is pulled out or do i have one that the wide flange has the flanges torn loose really don't i haven't seen one of those see you can tell where the old plate was here was the old plate right here see those lines and obviously they marked those you know with the plate so they could kind of measure how far the plate was distorted or something like that and this is both of these are rupture on net tension area to get rupture on the side you'd have to use fewer bolts otherwise you just really not going to get that rupture on the side and then this thing here still be connected thank you okay yeah every one of them there's a lot of work goes on in that really is yep uh-huh you too